| I'll use i instead of j for sqrt(-1) (since we lean more toward
being mathematicians than physicists in this conference).
Consider the general problem, x^y. Write y as c+di. Write x as a+bi.
Select O so that tan O = b/a and cos O has the same sign as a and sin O
has the same sign as b. (If a is zero, let O be an odd multiple of
pi/2.) Let r = a / cos O. (If a is zero, let r = b.) Then x = r cos
O + ir sin O. Let k = ln r. Note that the natural logarithm of x is
k+Oi because e^(k+Oi) = e^k * e^(Oi) = r * (cos O + i sin O).
Let z = x^y. Take the natural logarithm of both sides:
ln z = ln (x^y).
ln z = y * ln x.
ln z = y * (k + Oi).
ln z = (c+di)(k+Oi).
ln z = (ck-dO)+i(dk+cO).
Put both sides as powers of e:
z = e^[(ck-dO)+i(dk+CO)].
z = e^(ck-dO) + [cos (dk+cO) + i sin (dk+cO)].
For y = i, y = c+di = 0+1i, so c is 0 and d is 1. The above reduces
to:
z = e^(-O) + [cos k + i sin k].
For x = i, x = a+bi = 0+1i, so a is 0 and b is 1. Tan O = b/a,
so O is an odd multiple of pi/2. Since b is positive, we can let
O be pi/2, since that makes sin O positive also. Since a is 0,
r is b, which is 1. Since k = ln r, k = 0. Now the equation reduces
to:
z = e^(-pi/2) + [cos 0 + i sin 0].
z = e^(-pi/2).
It is real. It should be noted that there are an infinite number
of solutions, because any value can be chosen for O as long as it
has sin O = 1 and cos O = 0, such as 3pi/2. However, all solutions
for this problem are real.
-- edp
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